應用代數學 簡介應用代數學 簡介

2011. Sep. 2011. Sep.

李安莉 於輔大數學系李安莉 於輔大數學系

近世代數學近世代數學 近世代數是相對於古典代數來說的近世代數是相對於古典代數來說的 !! 古典代數基本上就是古典代數基本上就是方程論方程論 ,, 以探討方程以探討方程式的解為中心的式的解為中心的 ..

隨方程論的進展隨方程論的進展 , , 複數系的發現使得方程複數系的發現使得方程式求解的問題跨進了 一大步式求解的問題跨進了 一大步 ! ! 而一元方而一元方程式的根式解之研究引導出程式的根式解之研究引導出群群的概念的概念 ;; 漸漸漸的漸的 , , 方程式論不再是代數學的全部方程式論不再是代數學的全部 ,, 它它漸漸地轉向對代數結構的本身研究漸漸地轉向對代數結構的本身研究 !!

方程式論

高中 :

1. 代數基本定理

2. 根與係數的關係

3. Lagrange 內插公式

4. Newton 近似根求法

5. Strum 法則 ( 實係數多項式之根的分布情形 )

…

大學 :

Galois 理論

Q1: 給一元方程式 , 有無解存在 ?

Q2: 有無根式解存在 ?

Galois

群論

在多數場合下 , 代數理論指研究代數運算本身之性質 , 而不管施行對象之具體屬性…這種新觀點是 20 世紀才產生的 , 德國女數學家 Noether 在這方面有相當大的貢獻 .

AlgebraAlgebra 代數一詞 來自阿拉伯文代數一詞 來自阿拉伯文

al – jabral – jabr

通常的解釋是類似隱含在翻譯上面。這個通常的解釋是類似隱含在翻譯上面。這個詞詞的 的 al – jabr al – jabr 大概意味著什麼“復辟”或大概意味著什麼“復辟”或

“ “ 完成”，似乎是指移位減去條件對方的完成”，似乎是指移位減去條件對方的 方程 方程 ;;

關於 這學年的代數學關於 這學年的代數學 上課時間上課時間 : : 每週二第每週二第 3, 43, 4 堂堂 上課地點上課地點 : MA301 : MA301 評量方式評量方式 : : 期中考 期中考 30%30% 期末考 期末考 30%30% 平時考 平時考 40% 40%

教學網站教學網站 ::http://www.math.fju.edu.tw/blog/index.php?PID=44http://www.math.fju.edu.tw/blog/index.php?PID=44

Part 1. Part 1. Integers and Equivalence RelationsIntegers and Equivalence Relations

Chapter 0. Preliminaries:Chapter 0. Preliminaries:

1. Proofs1. Proofs

2. Sets2. Sets

3. Mappings3. Mappings

4. Equivalences4. Equivalences

◎◎ 等同學拿到課本後等同學拿到課本後 ,, 請自行復習之請自行復習之 !!

4. Equivalence Relations4. Equivalence Relations Def.Def. An equivalence relation on a set S is a set An equivalence relation on a set S is a set

R of ordered pairs of elements of S such that R of ordered pairs of elements of S such that

1. (a,a) lies in R for all a in S (reflexive)1. (a,a) lies in R for all a in S (reflexive)

2. (a,b) lies in R implies (b,a) lies in R ( symmetric)2. (a,b) lies in R implies (b,a) lies in R ( symmetric) 3. (a,b) lies in R and (b,c) lies in R implies (a,c) lies in R 3. (a,b) lies in R and (b,c) lies in R implies (a,c) lies in R

(transitive) (transitive)

Equivalence Classes Partition:The equivalence classes of an equivalence relation on a set S constitute a partition of S( a collection of nonempty disjoint subsets of S whose union is S)

For every partition P of S, there is an equivalence relation on S whose equivalence classes are the elements of P.

Chapter 1. Integers and permutationsChapter 1. Integers and permutations

1.1.1.1. Induction ( Induction ( 略略 )) 1.2.1.2. Divisors and Prime Factorization Divisors and Prime Factorization 1.3.1.3. Integers Modulo n Integers Modulo n 1.4.1.4. Permutations Permutations 1.5.1.5. Application to Cryptography Application to Cryptography

Sec.1.1. InductionSec.1.1. Induction Well-Ordering AxiomWell-Ordering Axiom: : Every nonempty set of nonnegative integeEvery nonempty set of nonnegative intege

rs has a smallest member.rs has a smallest member. The Principle of InductionThe Principle of Induction: If m is any integer. Let P(m), P(m+: If m is any integer. Let P(m), P(m+

1),…be statements such that 1),…be statements such that (1). P(m) is true,(1). P(m) is true, (2). P(k) (2). P(k) P(k+1) for all k P(k+1) for all k m m Then P(n) is true for all n Then P(n) is true for all n m. m.

Strong InductionStrong Induction: : If m is any integer. Let P(m), P(m+1),…be sIf m is any integer. Let P(m), P(m+1),…be statements such that tatements such that

(1). P(m) is true,(1). P(m) is true,

(2).If k (2).If k m and P(m),P(m+1),…,P(k) are true m and P(m),P(m+1),…,P(k) are true P(k+1) is also true P(k+1) is also true Then P(n) is true for all n Then P(n) is true for all n m. m.

Well-Ordering Induction Strong Induction Exercises!

1. Properties of Integers1. Properties of Integers Well Ordering PrincipleWell Ordering Principle: : Every nonempty set of inEvery nonempty set of in

tegers contains a smallest member. tegers contains a smallest member.

Division AlgorithmDivision Algorithm: : Let a and b be integers with b>Let a and b be integers with b>0. Then there exist unique integers q and r with the prope0. Then there exist unique integers q and r with the property that a=brty that a=b··q+r, where 0q+r, where 0≤r<b.≤r<b.

GCD:GCD: LCM:LCM: Fundamental Theorem of ArithmeticFundamental Theorem of Arithmetic: : Every integer n >1 is a prime or a product of primes.Every integer n >1 is a prime or a product of primes.

This product is uniquely determined up to the ordersThis product is uniquely determined up to the orders..

2. Modular Arithmetic2. Modular Arithmetic NotationNotation: a=q: a=q··b+r iff b| a-r iff b+r iff b| a-r iff a≡r mod b.a≡r mod b.

Some Basic PropertiesSome Basic Properties: Let n >0.: Let n >0. 1. a ≡ a mod n1. a ≡ a mod n 2. If a ≡ b mod n, then b ≡ a mod n2. If a ≡ b mod n, then b ≡ a mod n 3. If a ≡ b mod n and b ≡ c mod n, then a ≡ c mod n3. If a ≡ b mod n and b ≡ c mod n, then a ≡ c mod n 4. If a ≡ b mod n and c ≡ d mod n ,4. If a ≡ b mod n and c ≡ d mod n , then a+c ≡ b+d mod n and ac ≡ bd mod nthen a+c ≡ b+d mod n and ac ≡ bd mod n 5. If a ≡ b mod n, then a+c ≡ b+c mod n 5. If a ≡ b mod n, then a+c ≡ b+c mod n and ac ≡ bc mod n.and ac ≡ bc mod n.

Part 2.Views From Some Part 2.Views From Some ProblemsProblems

Geometric ConstructionsGeometric Constructions The ancient Greeks were fond of geometric constructions.The ancient Greeks were fond of geometric constructions. They were especially interested in constructions that could They were especially interested in constructions that could

be achieved using only be achieved using only a straightedgea straightedge without markings without markings and and a compassa compass..

Questions:Questions: 1. Can you bisect any given angel?1. Can you bisect any given angel? 2. Can you trisect any given angel?2. Can you trisect any given angel? 3. For any given positive integer n, can you construct a 3. For any given positive integer n, can you construct a regular polygon of n sides?regular polygon of n sides? 4. Given a circle, is it possible to construct a square with 4. Given a circle, is it possible to construct a square with same area?same area? 5. Given a side of a cube, is it possible to construct a cube5. Given a side of a cube, is it possible to construct a cube which has double the volume of the original cube?which has double the volume of the original cube?

Q1. Construction: Bisect AngleQ1. Construction: Bisect Angle Step 1. Draw an arc that is centered at the vertex of the angle. This arc can Step 1. Draw an arc that is centered at the vertex of the angle. This arc can

have a radius of any length. However, it must intersect both sides of the have a radius of any length. However, it must intersect both sides of the angle. We will call these intersection points angle. We will call these intersection points PP and and QQ This provides a point This provides a point on each line that is an equal distance from the vertex of the angle. on each line that is an equal distance from the vertex of the angle.

Step 2. Draw two more arcs. The first arc must be centered on one of the Step 2. Draw two more arcs. The first arc must be centered on one of the two points two points PP or or QQ. It can have any length radius. The second arc must be . It can have any length radius. The second arc must be centered on whichever point (P or Q) you did NOT choose for the first arc. centered on whichever point (P or Q) you did NOT choose for the first arc. The radius for the second arc MUST be the same as the first arc. Make sure The radius for the second arc MUST be the same as the first arc. Make sure you make the arcs long enough so that these two arcs intersect in at least you make the arcs long enough so that these two arcs intersect in at least one point. We will call this intersection point one point. We will call this intersection point XX. Every intersection point . Every intersection point between these arcs (there can be at most 2) will lie on the angle bisector. between these arcs (there can be at most 2) will lie on the angle bisector.

Step 3. Draw a line that contains both the vertex and Step 3. Draw a line that contains both the vertex and XX. Since the . Since the intersection points and the vertex all lie on the angle bisector, we know that intersection points and the vertex all lie on the angle bisector, we know that the line which passes through these points the line which passes through these points mustmust be the angle bisector. be the angle bisector.

Now, try to do this construction yourself. Now, try to do this construction yourself.

We can use We can use field theoryfield theory to prove it to prove it::

In 1801, Gauss asserted that a regular In 1801, Gauss asserted that a regular polygon of n sides is constructible if and only polygon of n sides is constructible if and only if n has the form if n has the form 22k k P P11PP2…2…PPss, where Pi’s are , where Pi’s are distinct primes of the form distinct primes of the form 22tt

+1.+1.

Example:Example:

regular polygonsregular polygons with 3,4,5,6,8,10,12,15,16,17 and with 3,4,5,6,8,10,12,15,16,17 and 20 sides are constructible. But regular polygons 20 sides are constructible. But regular polygons with 7,9,11,13,14,18,19 sides are not.with 7,9,11,13,14,18,19 sides are not.

22

One Example In Chemistry:One Example In Chemistry:

A benzene (A benzene ( 苯苯 ) molecule can be modeled as si) molecule can be modeled as six carbon arranged in a regular hexagon in a planx carbon arranged in a regular hexagon in a plane. At each carbon atom, one of three radicals Ne. At each carbon atom, one of three radicals NHH22, COOH, or OH can be attached. How many s, COOH, or OH can be attached. How many such compounds are possible?uch compounds are possible?

We can use Burnside’s theoremWe can use Burnside’s theorem

(one theorem coming from group action)(one theorem coming from group action)

to answer it.to answer it. There are also some interesting problems There are also some interesting problems

for coloring or counting.for coloring or counting.

Some Examples of Some Examples of GroupsGroups

Solvability of Polynomials by Radicals:Solvability of Polynomials by Radicals:

The equation The equation

The formulas for general cubic and quartic (fourth-degree polynomiaThe formulas for general cubic and quartic (fourth-degree polynomial) are more complicated, but they can be given in terms of radiacls ol) are more complicated, but they can be given in terms of radiacls of rational expressions of the coefficients.f rational expressions of the coefficients.

Is it solvable by radicals for a quintic ( fifth-degree polynomial)? Is it solvable by radicals for a quintic ( fifth-degree polynomial)? Use Use Galois theoryGalois theory we can prove that 3 we can prove that 3xx5 5 - 15x+5 is not solvable by r - 15x+5 is not solvable by r

adicals.adicals.